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Ksenya-84 [330]
2 years ago
12

A 1.50-liter sample of dry air in a cylinder exerts a pressure of 3.00 atmospheres at a temperature of 25°C. Without changing th

e temperature, a piston is moved in the cylinder until the pressure in the cylinder is reduced to 1.00 atmospheres. The volume of the gas is _____.

Chemistry
2 answers:
Aneli [31]2 years ago
3 0
Hope this helps you.

Galina-37 [17]2 years ago
3 0

Answer:

V_2=4.5 L

Explanation:

Hi, air can be considered as an ideal gas. If mantaining constant the temeperature, you can apply the Boyle-Mariotte formula:

P_1*V_1=P_2*V_2

In this case:

P_1=3 atm

V_1=1.5 L

P_2=1 atm

So:

V_2=\frac{P_1*V_1}{P_2}

V_2=\frac{3atm*1.5 L}{1atm}

V_2=4.5 L

You might be interested in
If you start with 100 grams of hydrogen-3, how many grams will you have after 24.6 years?
svet-max [94.6K]

Answer:

The mass left after 24.6 years is 25.0563 grams

Explanation:

The given parameters are;

The mass of the hydrogen-3 = 100 grams

The half life of hydrogen-3 which is also known as = 12.32 years

The formula for calculating half-life is given as follows;

N(t) = N_0 \times \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{\frac{1}{2} }} }

Where;

N(t) = The mass left after t years

N₀ =  The initial mass of the hydrogen-3 = 100 g

t = Time duration of the decay = 24.6 years

t_{\frac{1}{2} } = Half-life = 12.32 years

N(24.6) = 100 \times \left (\dfrac{1}{2} \right )^{\dfrac{24.6}{12.32}} } = 25.0563

The mass left after 24.6 years = 25.0563 grams.

4 0
2 years ago
Carol and Juan were playing with Carol's new train set. They pushed the trains around the track. They tried four different
Bumek [7]

Speed (m/s) = distance (metres) ÷ time (seconds)

For Track 1: 0.2 ÷ 2 = 0.1m/s

For Track 2: 0.2 ÷ 2 = 0.1m/s

For Track 3: 0.6 ÷ 6 = 0.1m/s

For Track 4: 0.4 ÷ 6 = 0.07m/s

The train on Track 4 had the slowest speed because it's got the shortest speed and it's covering less distance per second therefore it is slower.

6 0
2 years ago
Even though we are well into the course Dumbledore notices some of the students are still having trouble with significant figure
Sphinxa [80]

Answer:

B) 2

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

The rule apply for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

The rule apply for the addition and subtraction is :

The least precise number present after the decimal point determines the number of significant figures in the answer.

(3.478-2.31) = 1.168 ≅ 1.17 (Rounded to least decimal digit)

(4.428-3.56) = 0.868 ≅ 0.87 (Rounded to least decimal digit)

So,

1.17 * 0.87 = 1.0  (Rounded to least significant)

Answer - two significant digits

5 0
2 years ago
A patient is receiving 3000 mL/day of a solution that con-
frutty [35]

Answer:

600 kCal/ day

Explanation:

You can solve this type of problems using proportions, with the rule of three.

Follow these steps:

1. Analyze the giving data

2. Organize it

3.Write the proportions  

4. Clear the value you are looking for

5. Solve the equations by multiplying or dividing

Now for the exercise:

You have the following data:

3000 mL Solution per 1 day ----> 3000 mL S

The concentration of the solution is  

5 g of dextrose per 100 mL  Solution ----> 5g D - 100 mL S

The glucose provides some specific energy

4 kCal per 1 g of dextrose ----> 4kCal - 1 g D

First find the total grams of  dextrose (glucose) given in one day.

3000 mL S - x g D?

 100 mL S - 5 g D

The grams of glucose given in one day

xg D? = 3000 mL S(5 g D)/100 mL S  

you also can cancel the units mL S

xg D? = 150 g D.

Answer: The grams given in one day of glucose are 150 g /day

Now find the total Energy in those grams:

4 kCal -      1 g D

x kCal?-150 g D

Clear x kCal?

x kCal? = 4kCal (150g D)/(1g D)

x kCal? = 600 kCal  remember is for one day!

so the answer is

The patient is receiving 600 kCal/day

5 0
2 years ago
A solution contains 4.08 g of chloroform (CHCl3) and 9.29 g of acetone (CH3COCH3). The vapor pressures at 35 ∘C of pure chlorofo
Sergeeva-Olga [200]

Explanation:

Below is an attachment containing the solution.

8 0
2 years ago
Read 2 more answers
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